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Research Methodology - Dr. Krishan K. Pandey

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Sampling Fundamentals 171<br />

08 .<br />

= ×<br />

64<br />

08 .<br />

= ×<br />

64<br />

2400 − 64<br />

2400 − 1<br />

2336<br />

2399<br />

= (0.1) (.97)<br />

= .097<br />

(3) 90 per cent confidence interval for the mean number of accidents per intersection per year<br />

is as follows:<br />

R<br />

S|<br />

T|<br />

σ s X ± z ×<br />

n<br />

N − n<br />

N − 1<br />

b gb g<br />

= 3. 2 ± 1645 . . 097<br />

U<br />

V|<br />

W|<br />

= 32 . ± . 16 accidents per intersection.<br />

When the sample size happens to be a large one or when the population standard deviation is<br />

known, we use normal distribution for detemining confidence intervals for population mean as stated<br />

above. But how to handle estimation problem when population standard deviation is not known and<br />

the sample size is small (i.e., when n < 30 )? In such a situation, normal distribution is not appropriate,<br />

but we can use t-distribution for our purpose. While using t-distribution, we assume that population is<br />

normal or approximately normal. There is a different t-distribution for each of the possible degrees of<br />

freedom. When we use t-distribution for estimating a population mean, we work out the degrees of<br />

freedom as equal to n – 1, where n means the size of the sample and then can look for cirtical value<br />

of ‘t’ in the t-distribution table for appropriate degrees of freedom at a given level of significance. Let<br />

us illustrate this by taking an example.<br />

Illustration 3<br />

The foreman of ABC mining company has estimated the average quantity of iron ore extracted to be<br />

36.8 tons per shift and the sample standard deviation to be 2.8 tons per shift, based upon a random<br />

selection of 4 shifts. Construct a 90 per cent confidence interval around this estimate.<br />

Solution: As the standard deviation of population is not known and the size of the sample is small, we<br />

shall use t-distribution for finding the required confidence interval about the population mean. The<br />

given information can be written as under:<br />

X = 368 . tons per shift<br />

σ s = 28 . tons per shift<br />

n = 4<br />

degrees of freedom = n – 1 = 4 – 1 = 3 and the critical value of ‘t’ for 90 per cent confidence interval<br />

or at 10 per cent level of significance is 2.353 for 3 d.f. (as per the table of t-distribution).

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